hoffman and kunze solution
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hoffman and kunze solution

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Linear Algebra
Hoffman & Kunze
2nd edition
Answers and Solutions to Problems and Exercises
Typos, comments and etc...
Gregory R. Grant
University of Pennsylvania
email: ggrant@upenn.edu
Julyl 2017
This is one of the ultimate classic textbooks in mathematics. It will probably be read for generations to come. Yet I
cannot find a comprehensive set of solutions online. Nor can I find lists of typos. Since I\u2019m going through this book in some
detail I figured I\u2019d commit some of my thoughts and solutions to the public domain so others may have an easier time than I
have finding supporting matericals for this book. Any book this classic should have such supporting materials readily avaiable.
If you find any mistakes in these notes, please do let me know at one of these email addresses:
ggrant543@gmail.com or greg@grant.org or ggrant@upenn.edu
The latex source file for this document is available at http://greg.grant.org/hoffman and kunze.tex. Use it wisely.
And if you cannot manage to download the link because you included the period at the end of the sentence, then maybe you
are not smart enough to be reading this book in the first place.
Chapter 1: Linear Equations
Section 1.1: Fields
Page 3. Hoffman and Kunze comment that the term \u201ccharacteristic zero\u201d is \u201cstrange.\u201d But the characteristic is the smallest n
such that n · 1 = 0. In a characteristic zero field the smallest such n is 0. This must be why they use the term \u201ccharacteristic
zero\u201d and it doesn\u2019t seem that strange.
Section 1.2: Systems of Linear Equations
Page 5 Clarification: In Exercise 6 of this section they ask us to show, in the special case of two equations and two unknowns,
that two homogeneous linear systems have the exact same solutions then they have the same row-reduced echelon form (we
know the converse is always true by Theorem 3, page 7). Later in Exercise 10 of section 1.4 they ask us to prove it when
there are two equations and three unknowns. But they never tell us whether this is true in general (for abitrary numbers of
unknowns and equations). In fact is is true in general. This explanation was given on math.stackexchange:
Solutions to the homogeneous system associated with a matrix is the same as determining the null space of the
relevant matrix. The row space of a matrix is complementary to the null space. This is true not only for inner
product spaces, and can be proved using the theory of non-degenerate symmetric bilinear forms.
So if two matrices of the same order have exactly the same null space, they must also have exactly the same row
space. In the row reduced echelon form the nonzero rows form a basis for the row space of the original matrix,
and hence two matrices with the same row space will have the same row reduced echelon form.
Exercise 1: Verify that the set of complex numbers described in Example 4 is a subfield of C.
Solution: Let F = {x + y\u221a2 | x, y \u2208 Q}. Then we must show six things:
1. 0 is in F
2. 1 is in F
3. If x and y are in F then so is x + y
4. If x is in F then so is \u2212x
5. If x and y are in F then so is xy
6. If x , 0 is in F then so is x\u22121
For 1, take x = y = 0. For 2, take x = 1, y = 0. For 3, suppose x = a+b
2 and y = c+d
2. Then x+y = (a+c)+(b+d)
2 \u2208 F.
For 4, suppose x = a + b
2. Then \u2212x = (\u2212a) + (\u2212b)\u221a2 \u2208 F. For 5, suppose x = a + b\u221a2 and y = c + d\u221a2. Then
2 Chapter 1: Linear Equations
xy = (a + b
2)(c + d
2) = (ac + 2bd) + (ad + bc)
2 \u2208 F. For 6, suppose x = a + b\u221a2 where at least one of a or b is not
zero. Let n = a2 \u2212 2b2. Let y = a/n + (\u2212b/n)\u221a2 \u2208 F. Then xy = 1n (a + b
2)(a \u2212 b\u221a2) = 1n (a2 \u2212 2b2) = 1. Thus y = x\u22121 and
y \u2208 F.
Exercise 2: Let F be the field of complex numbers. Are the following two systems of linear equations equivalent? If so,
express each equation in each system as a linear combination of the equations in the other system.
x1 \u2212 x2 = 0 3x1 + x2 = 0
2x1 + x2 = 0 x1 + x2 = 0
Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the
second, and conversely.
3x1 + x2 =
(x1 \u2212 x2) + 43(2x1 + x2)
x1 + x2 =
(x1 \u2212 x2) + 23(2x1 + x2)
x1 \u2212 x2 = (3x1 + x2) \u2212 2(x1 + x2)
2x1 + x2 =
(3x1 + x2) +
(x1 + x2)
Exercise 3: Test the following systems of equations as in Exercise 2.
\u2212x1 + x2 +4x3 = 0 x1 \u2212 x3= 0
x1 + 3x2+8x3 = 0 x2 + x3= 0
2 x1 + x2 + 52 x3 = 0
Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the
second, and conversely.
x1 \u2212 x3 = \u221234 (\u2212x1 + x2 + 4x3) + 14 (x1 + 3x3 + 8x3)
x2 + 3x3 = 14 (\u2212x1 + x2 + 4x3) + 14 (x1 + 3x3 + 8x3)
\u2212x1 + x2 + 4x3 = \u2212(x1 \u2212 x3) + (x2 + 3x3)
x1 + 3x2 + 8x3 = (x1 \u2212 x3) + 3(x2 + 3x3)
2 x1 + x2 + 52 x3 = 12 (x1 \u2212 x3) + (x2 + 3x3)
Exercise 4: Test the following systems as in Exercie 2.
2x1 + (\u22121 + i)x2 + x4 = 0 (1 + i2 )x1+8x2 \u2212 ix3 \u2212x4 = 0
3x2 \u2212 2ix3 + 5x4 = 0 23 x1 \u2212 12 x2 + x3 +7x4 = 0
Solution: These systems are not equivalent. Call the two equations in the first system E1 and E2 and the equations in the
second system E\u20321 and E
2. Then if E
2 = aE1 + bE2 since E2 does not have x1 we must have a = 1/3. But then to get the
coefficient of x4 we\u2019d need 7x4 = 13 x4 + 5bx4. That forces b =
3 . But if a =
3 and b =
3 then the coefficient of x3 would have
to be \u22122i 43 which does not equal 1. Therefore the systems cannot be equivalent.
Exercise 5: Let F be a set which contains exactly two elements, 0 and 1. Define an addition and multiplication by the tables:
+ 0 1
0 0 1
1 1 0
· 0 1
0 0 0
0 0 1
Section 1.2: Systems of Linear Equations 3
Solution: We must check the nine conditions on pages 1-2:
1. An operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right.
This is true for the addition table so addition is commutative.
2. There are eight cases. But if x = y = z = 0 or x = y = z = 1 then it is obvious. So there are six non-trivial cases. If there\u2019s
exactly one 1 and two 0\u2019s then both sides equal 1. If there are exactly two 1\u2019s and one 0 then both sides equal 0. So addition
is associative.
3. By inspection of the addition table, the element called 0 indeed acts like a zero, it has no effect when added to another
4. 1 + 1 = 0 so the additive inverse of 1 is 1. And 0 + 0 = 0 so the additive inverse of 0 is 0. In other words \u22121 = 1 and
\u22120 = 0. So every element has an additive inverse.
5. As stated in 1, an operation is commutative if the table is symmetric across the diagonal that goes from the top left to the
bottom right. This is true for the multiplication table so multiplication is commutative.
6. As with addition, there are eight cases. If x = y = z = 1 then it is obvious. Otherwise at least one of x, y or z must equal 0.
In this case both x(yz) and (xy)z equal zero. Thus multiplication is associative.
7. By inspection of the multiplication table, the element called 1 indeed acts like a one, it has no effect when multiplied to
another element.
8. There is only one non-zero element, 1. And 1 · 1 = 1. So 1 has a multiplicative inverse. In other words 1\u22121 = 1.
9. There are eight cases. If x = 0 then clearly both sides equal zero. That takes care of four cases. If all three x = y = z = 1
then it is obvious. So we are down to three cases. If x = 1 and y = z = 0 then both sides are zero. So we\u2019re down to the two
cases where x = 1 and one of y or z equals 1 and the other equals 0. In this case both sides equal 1. So x(y + z) = (x + y)z in
all eight cases.
Exercise 6: Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they
are equivalent.
Solution: Write the two systems as follows:
a11x + a12y = 0
a21x + a22y = 0
am1x + am2y = 0
b11x + b12y = 0
b21x + b22y